Question: Evaluate the triple integral. $ \int_{-1}^1 \int_0^{y^2} \int_{-x}^x 3z^2 + x - y \, dz \, dx \, dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{19}{63}$ (Choice B) B $\dfrac{13}{63}$ (Choice C) C $\dfrac{11}{63}$ (Choice D) D $\dfrac{17}{63}$
Solution: We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_{-1}^1 \int_0^{y^2} \int_{-x}^x 3z^2 + x - y \, dz \, dx \, dy \\ \\ &=\int_{-1}^1 \int_0^{y^2} \left[ z^3 + xz - yz \right]_{-x}^x dx \, dy \\ \\ &=\int_{-1}^1 \int_0^{y^2} \left( x^3 + x^2 - xy \right) - \left( -x^3 - x^2 + xy \right) dx \, dy \\ \\ &=\int_{-1}^1 \int_0^{y^2} 2x^3 + 2x^2 - 2xy \, dx \, dy \end{aligned}$ The second layer: $\begin{aligned} &\int_{-1}^1 \int_0^{y^2} 2x^3 + 2x^2 - 2xy \, dx \, dy \\ \\ &= \int_{-1}^1 \left[ \dfrac{x^4}{2} + \dfrac{2x^3}{3} - x^2y \right]_0^{y^2} dy \\ \\ &= \int_{-1}^1 \dfrac{y^8}{2} + \dfrac{2y^6}{3} - y^5 \, dx \end{aligned}$ The third layer: $\begin{aligned} &\int_{-1}^1 \dfrac{y^8}{2} + \dfrac{2y^6}{3} - y^5 \, dx \\ \\ &= \left[ \dfrac{y^9}{18} + \dfrac{2y^7}{21} - \dfrac{y^6}{6} \right]_{-1}^1 \\ \\ &= \left( \dfrac{1}{18} + \dfrac{2}{21} - \dfrac{1}{6} \right) - \left( \dfrac{-1}{18} - \dfrac{2}{21} - \dfrac{1}{6} \right) \\ \\ &= \dfrac{19}{63} \end{aligned}$ In conclusion: $ \int_{-1}^1 \int_0^{y^2} \int_{-x}^x 3z^2 + x - y \, dz \, dx \, dy = \dfrac{19}{63}$